User:Fropuff/Drafts/Unit quaternion group

''This is a draft article on the group of unit quaternions. I am not sure what the best name is for such an article. Suggestions are welcome. Others are welcome to contribute to this page; but, as it is still in my user space, I reserve the right to mutilate any contributions.''

Suggested names:
 * Sp(1)
 * SU(2)
 * Spin(3)
 * 3-sphere group
 * unit quaternion group
 * group of unit quaternions

The unit quaternion group is the set of all unit quaternions, i.e. those quaternions with absolute value equal to 1. As the name implies, they form a group under multiplication&mdash;a subgroup of the multiplicative group of all non-zero quaternions. This group is sometimes denoted Sp(1) since it is the first in the family of (compact) symplectic groups. In symbols,
 * $$\mathrm{Sp}(1) = \{q \in \mathbb H^\times : |q| = 1\}.$$

Some other common notations for the unit quaternion group include U(1,H) and SL(1,H).

The unit quaternion group can be understood as the quaternionic analog of the circle group&mdash;the set of unit complex numbers under complex multiplication. Rather than a circle, the set of unit quaternions make up a unit 3-sphere inside H since the quaternions are 4-dimensional. Algebraically, the primary difference with the circle group is that the unit quaternion group is nonabelian. This follows from the fact that quaternion multiplication is noncommutative.

The group Sp(1) is the primary, and perhaps most important, example of a Lie group; which is to say that the group space is a manifold and the group operations (multiplication and inversion) are smooth. It is, in fact, the simplest example of a compact, simply-connected, nonabelian Lie group.

The unit quaternion group appears in a variety of forms in mathematics. The most common alternative form is as the special unitary group SU(2), the group of all 2&times;2 unitary matrices with unit determinant. It also appears as the spin group Spin(3) which is a double cover of the rotation group SO(3). These isomorphisms and others are explained below.

Group structure
To verify that the unit quaternions form a subgroup of H&times; one needs to check that they are closed under multiplication and that every element has an inverse which is also a unit quaternion. These properties follow from the relations
 * $$|qp| = |q||p|\qquad\mbox{and}\qquad|q^{-1}| = \frac{1}{|q|}$$

which hold for all p, q &isin; H&times;.

The inverse of a unit quaternion q is given by the quaternionic conjugate
 * $$q^{-1} = \bar q$$

since
 * $$\bar q q = q\bar q = |q|^2 = 1.$$

A quaternion has absolute value 1 if and only if its inverse is equal to its conjugate.

Just as for complex numbers, every nonzero quaternion q can be written uniquely as a product of a positive real number (|q|) and a unit quaternion (q/|q|). In fact, the multiplicative group H&times; is isomorphic to the direct product of Sp(1) and the multiplicative group of positive real numbers.
 * $$\mathbb H^\times = \mathbb R^{+} \times \mathrm{Sp}(1).$$

As a Lie group
Sp(1) has a natural Lie group structure, with a topology and smooth structure inherited from H &cong; R4. Since the group manifold of Sp(1) is a 3-sphere it follows that Sp(1) is a compact, connected, and simply connected Lie group of dimension 3. Indeed, Sp(1) is the simplest example of a compact nonabelian Lie group. It is of fundamental importance in the theory of Lie groups.

As a symplectic group
The notation Sp(1) for the group of unit quaternions comes from the fact that Sp(1) is the first in the family of compact symplectic groups, denoted Sp(n). The quaternionic linear group GL(n,H) is the group of all invertible n&times;n quaternionic matrices under matrix multiplication. The group Sp(n) is the subgroup which preserves the standard hermitian form on Hn (thought of as a right H-vector space):
 * $$\langle x, y\rangle = \bar x_1 y_1 + \cdots + \bar x_n y_n$$

The group GL(1,H) is just the multiplicative group on nonzero quaternions H&times;, which acts on H by left multiplication. The group Sp(1) is the subgroup of quaternions q for which
 * $$\langle qx, qy\rangle = \overline{qx}qy = \bar x\bar qqy = \bar xy = \langle x, y\rangle.$$

for all x,y &isin; H. This is precisely the group of unit quaternions:
 * $$\{q\in \mathbb H^\times : \bar qq = |q|^2 = 1\}.$$

The compact symplectic groups are also sometimes called hyperunitary groups and denoted U(n,H), so that Sp(1) = U(1,H).

Pure unit quaternions
The pure or imaginary quaternions, denoted by Im H, are those quaternions whose real part is 0. The pure unit quaternions are the pure quaternions with absolute value 1. These are quaternions of the form
 * $$\tau = x_1 i + x_2 j + x_3 k\;\mbox{ where }\;x_1^2 + x_2^2 + x_3^2 = 1.$$

Geometrically, the pure units quaternions form a 2-sphere making up the "equator" of the unit 3-sphere in H (with &plusmn;1 as the poles). They form a single conjugacy class in Sp(1).

One can show that a quaternion $$\tau$$ is a pure unit quaternion if and only if $$\tau^2 = -1$$.

Exponential map and one-parameter subgroups
Just as every element of the circle group can be written in exponential form $$e^{i\theta}$$, every unit quaternion can be written as $$e^{\tau\psi}$$ where $$\tau$$ is a pure unit quaternion and $$\psi$$ is a real number.

The exponential map
The exponential function on the quaternions can be defined just as it is on the complex numbers, i.e. by its power series expansion:
 * $$e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}.$$

This series converges for all x &isin; H.

In the complex case, one obtains a map to the circle group by restricting the exponential function to the imaginary numbers. For the quaternions, one obtains a map to Sp(1) by restricting to the imaginary quaternions:
 * $$\exp\colon \mathrm{Im}\,\mathbb H \to \mathrm{Sp}(1)$$

This follows from the fact that for x &isin; Im H we have
 * $$\overline{e^x} = e^{\bar x} = e^{-x} = \left(e^x\right)^{-1}.$$

Every unit quaternion can be written as the exponential of a pure quaternion so the above map is surjective. Using the formulas in the next section one can show that the exponential map is surjective onto Sp(1) when restricted to a ball of radius $$\pi$$ inside Im H. In fact, the mapping is bijective in this region except for the boundary of the ball, all of which maps to the point &minus;1 &isin; Sp(1). (This reflects the topological fact that the 3-sphere can be constructed by identifying the boundary of a 3-ball to a point).

Euler's formula
For any pure unit quaternion $$\tau$$ and real number $$\psi$$, the exponential of $$\tau\psi$$ is given by the quaternionic analog of Euler's formula:
 * $$e^{\tau\psi} = \cos\psi + \tau\,\sin\psi.$$

This follows purely from the definition of the exponential and the fact that $$\tau^2 = -1$$. The proof is just as in the complex case where $$\tau = i$$. Note that the quaternionic analog of Euler's identity
 * $$e^{\tau\pi} = -1\,$$

is valid for all $$\tau$$ with $$\tau^2 = -1$$.

Any pure quaternion $$x$$ can be written in the form $$\tau\psi$$ where $$\tau$$ is a pure unit quaternion and $$\psi$$ is a nonnegative real number. Simply, take $$\tau = x/|x|$$ and $$\psi = |x|$$. Euler's formula then gives a method for computing the exponential of any pure quaternion $$x$$.

One-parameter subgroups
For every pure unit quaternion $$\tau$$, we have a one-dimensional Lie subgroup of Sp(1) given by the set of points
 * $$\{e^{\tau\psi} : \psi \in \mathbb R\}$$

The group law is just
 * $$e^{\tau\psi_1}e^{\tau\psi_2} = e^{\tau(\psi_1 + \psi_2)}.\,$$

These groups are just copies of the circle group inside Sp(1). For each such $$\tau$$, we have a Lie group homomorphism from R to Sp(1). In fact, for all x &isin; Im H we have a Lie group homomorphism R &rarr; Sp(1) given by
 * $$\psi \mapsto e^{x\psi}$$

These maps are called one-parameter subgroups of Sp(1). Note carefully the distinction between one-parameter subgroups, which are given by homomorphism and have a fixed parametrization, and one-dimensional subgroups, which are just Lie subgroups without any specified parametrization.

Action on the quaternions
The group Sp(1) has a natural smooth action on the quaternions H given by conjugation. For each q in Sp(1) and x in H this action is given by
 * $$\rho_q(x) = qx\bar q.$$

The function $$\rho_q$$ is easily seen be an automorphism of H. Since $$\rho$$ is an action the map
 * $$\rho : \mathrm{Sp}(1) \to \mathrm{Aut}(\mathbb H)$$

which sends $$q$$ to $$\rho_q$$ is a homomorphism of Lie groups. In fact, we shall see that this map is a surjective homomorphism with kernel {&plusmn;1}.

One can show that every automorphism of H leaves the decomposition H = R &oplus; Im H invariant. That is to say, every automorphism of H reduces to an action on Im H. It follows that every automorphism of H preserves the norm on H, and is therefore a given by an orthogonal transformation. By examining the action on the basis i, j, k one can see that such a transformation must have determinant one. The automorphism group of H is therefore contained in the rotation group SO(3):


 * $$\mathrm{Aut}(\mathbb H) \sub \mathrm{SO}(\mathrm{Im}\,\mathbb H) = \mathrm{SO}(3).$$

Stated simply, every automorphism of H leaves the real part of a quaternion invariant and acts via a rotation on the imaginary part.

Since conjugation by a unit quaternion q is an automorphism it must act by rotation on Im H. In fact, one can show that conjugation by
 * $$q = e^{\tau\psi} = \cos\psi + \tau\sin\psi\,$$

(where $$\tau$$ is a pure unit quaternion) corresponds to a counterclockwise rotation about $$\tau$$ through an angle of $$2\psi$$. For details see the article on quaternions and spatial rotations.

Since every 3-dimensional rotation is a simple rotation about some axis, and every such rotation is given by conjugation by some unit quaternion, it follows that the automorphism group of H is all of SO(3) and the map $$\rho$$ is a surjection.

The map $$\rho$$ is not an isomorphism, however, since it is not injective. The kernel is the set of unit quaternions that commute with all of H. This is just the intersection of Sp(1) with the center of H. The center of H is just the real numbers R whose intersection with Sp(1) is {&plusmn;1}. Therefore the map $$\rho$$ is a 2:1 homomorphism. Conjugation by q and &minus;q give the same rotation.

It follows from the first isomorphism theorem that
 * $$\mathrm{Sp}(1)/\{\pm 1\} \cong \mathrm{Aut}(\mathbb H) = \mathrm{SO}(3).$$

Self action
Like all groups, Sp(1) acts on itself by conjugation. One simply restricts $$\rho_q$$ to Sp(1), giving a homomorphism
 * $$\rho : \mathrm{Sp}(1) \to \mathrm{Aut}(\mathrm{Sp}(1)).\,$$

The kernel of this homomorphism is just the center of Sp(1), which is again just {&plusmn;1}. The inner automorphism group of Sp(1) is therefore isomorphic to SO(3). In fact, there are no outer automorphisms so this is the whole automorphism group:
 * $$\mathrm{Aut}(\mathrm{Sp}(1)) \cong \mathrm{SO}(3).$$

The conjugation action of the unit quaternion group on itself is given by rotation of the pure part of each unit quaternion. Viewing Sp(1) as a 3-sphere with poles at &plusmn;1, this action is just rotation of the sphere about the polar axis. Since any rotation is possible, the conjugacy classes of Sp(1) are the 2-spheres at a fixed "latitude", i.e. with a given real part. There are two degenerate classes at the poles, with &plusmn;1 each belonging to a singleton class. The space of conjugacy classes is therefore diffeomorphic to the real interval [&minus;1,+1].

The centralizer of any unit quaternion is just the stabilizer under conjugation. For any unit quaternion q not lying in the center of Sp(1) the centralizer is the unique circle group containing q. It follows from the orbit-stabilizer theorem that the coset space Sp(1)/T (where T is any circle subgroup) is diffeomorphic to a 2-sphere. The group Sp(1) can then be viewed a principal circle bundle over the 2-sphere. This is the so-called Hopf bundle.

Adjoint representation
Every Lie group has a natural representation on its own Lie algebra called the adjoint representation. As explained below, the Lie algebra of Sp(1) can be identified with the space of pure quaternions, Im H. The adjoint action of Sp(1) on Im H is given simply by conjugation. That is, the adjoint map
 * $$\mathrm{Ad} : \mathrm{Sp(1)} \to \mathrm{Aut}(\mathrm{Im}\,\mathbb H)$$

is simply the restriction of $$\rho$$ to the pure quaternions. Indeed, the map $$\rho$$ is often called the adjoint map.

Automorphisms of a Lie algebra are linear maps with preserve the Lie bracket. So rotations are not only automorphisms of H (preserving quaternionic multiplication) but automorphisms of Im H (persevering the Lie bracket). The automorphism group of Im H is, again, the rotation group SO(3).

SU(2)
The algebra of quaternions has a convenient matrix representation over the complex numbers. Specifically, H is isomorphic to the real algebra of 2&times;2 complex matrices of the form
 * $$\begin{bmatrix}\alpha & -\bar \beta \\ \beta & \bar\alpha\end{bmatrix}.$$

One can check that the set of all such matrices forms a real subalgebra of M2(C). Quaternionic conjugation on this algebra is given by taking the conjugate transpose and the norm squared is given by taking the determinant.

In this representation, the group of unit quaternions is given by the set of all such matrices whose determinant is equal to 1. This group is precisely the special unitary group SU(2):
 * $$\mathrm{SU}(2) = \left\{\begin{bmatrix}\alpha & -\bar \beta \\ \beta & \bar\alpha\end{bmatrix} : \alpha^2 + \beta^2 = 1\right\}$$

The group SU(2) may be regarded as the intersection of the unitary group U(2) and the special linear group SL(2,C). In general, one can always represent n&times;n quaternionic matrices as 2n&times;2n complex matrices. The image of Sp(n) under such a representation is the intersection of U(2n) and the symplectic group Sp(2n,C) and is frequently denoted by USp(2n). For n = 1, one has Sp(2,C) = SL(2,C) so that USp(2) = SU(2).

Although the groups Sp(1) and SU(2) are isomorphic there is no canonical isomorphism between them. One must first choose a suitable identification of H with C2. For example, writing a quaternion q in the form &alpha; + j&beta; where &alpha; and &beta; are complex numbers gives an isomorphism
 * $$(\alpha + j\beta) \mapsto \begin{bmatrix}\alpha & -\bar\beta \\ \beta & \bar\alpha\end{bmatrix}.$$

This map preserves the norm so it maps Sp(1) onto SU(2). All other isomorphisms can be obtained from this one by first conjugating q.

Spin(3)
The group of unit quaternions has a natural action on R3 when identified with the space of pure quaternions, Im H. This action is given by
 * $$\gamma_q(x) = qx\bar q\,$$

for all q &isin; Sp(1) and all x &isin; Im H. This action amounts to a rotation of the vector x. Explicitly, if
 * $$q = e^{\tau\psi} = \cos\psi + \tau\sin\psi\,$$

for some pure unit quaternion &tau;, then the map $$\gamma_q$$ corresponds to a counterclockwise rotation about the unit vector $$\tau$$ through an angle of $$2\psi$$. The factor of 2 here means that both q and &minus;q map to the same rotation.

The map $$\gamma$$ then gives a surjective Lie group homomorphism from Sp(1) to the rotation group SO(3) with kernel {&plusmn;1}.
 * $$\gamma\colon \mathrm{Sp}(1) \to \mathrm{SO}(3).\,$$

Since Sp(1) is connected and simply connected it acts as the universal covering group of SO(3) with $$\gamma$$ as the covering homomorphism. In the theory of the orthogonal groups, the universal cover of SO(n) for n &ge; 3 is called a spin group and denoted by Spin(n). The group of unit quaternions is then isomorphic to the spin group Spin(3):
 * $$\mathrm{Sp}(1) \cong \mathrm{Spin}(3).$$

For more on the relationship between Sp(1) and SO(3) see the article: quaternions and spatial rotations.

There is another interpretation of the map &gamma; : Sp(1) &rarr; SO(3). If one thinks of Im H as the Lie algebra of Sp(1), then the map $$\gamma$$ is the adjoint action of Sp(1) on its own Lie algebra.

Embeddings in SO(4)
Quaternions act naturally on R4 = H by left and right multiplication. These actions are orthogonal when restricted to the group of unit quaternions. This gives rise to two embeddings of Sp(1) in SO(4) called the left and right embeddings.

For q in H let $$L_q$$ and $$R_q$$ denote the left and right multiplication maps respectively. That is,
 * $$L_q(p) = qp\qquad R_q(p) = pq$$

for all q and p in H. These maps satisfy the identites
 * $$\begin{align}

&L_pL_q = L_{pq}\qquad & &R_pR_q = R_{qp}\\ &L_{\bar q} = (L_q)^T & &R_{\bar q} = (R_q)^T\\ &\det(L_q) = |q|^2    & &\det(R_q) = |q|^2 \end{align}$$ It follows that the maps
 * $$q \mapsto L_q \qquad q \mapsto R_{\bar q}$$

are homomorphisms from Sp(1) to SO(4). They are embeddings since the kernel in each case is trivial. The images are sometimes denoted Sp(1)L and Sp(1)R.

Explicitly, for $$q = q_0 + q_1 i + q_2 j + q_3 k$$ one has
 * $$L_q = \begin{bmatrix}

q_0 & -q_1 & -q_2 & -q_3 \\ q_1 & q_0 & -q_3 & q_2 \\ q_2 & q_3 & q_0 & -q_1 \\ q_3 & -q_2 & q_1 & q_0 \end{bmatrix}\qquad R_{\bar q} = \begin{bmatrix} q_0 & q_1 & q_2 & q_3 \\ -q_1 & q_0 & -q_3 & q_2 \\ -q_2 & q_3 & q_0 & -q_1 \\ -q_3 & -q_2 & q_1 & q_0 \end{bmatrix}$$

Discrete subgroups
The discrete subgroups of Sp(1) fall into and ADE classification


 * An: cyclic groups
 * Dn: dicyclic groups
 * E6: binary tetrahedral group
 * E7: binary octahedral group
 * E8: binary icosahedral group

Lie algebra
The Lie algebra of a Lie group is defined as the tangent space to the identity element of the group together with a bilinear operator, called the Lie bracket, which is induced by group multiplication.

For the group Sp(1), the tangent space to the identity is the three-dimensional space of all pure quaternions, Im H. The Lie bracket is given by the commutator:
 * $$[x,y] = xy - yx.\,$$

A basis for Im H is given by the quaternions i, j, and k. These have the commutation relations:
 * $$[i,j] = 2k\,$$
 * $$[j,k] = 2i\,$$
 * $$[k,i] = 2j\,$$

and
 * $$[i,i]=[j,j]=[k,k]=0.\,$$

Representations
See: representation theory of SU(2)

Related topics

 * 3-sphere
 * quaternions and spatial rotation
 * rotation group
 * circle group
 * symplectic group

Category:Lie groups | Category:Lie algebras | Category:Quaternions